Question

The pH of a solution is 3.2. Calculate [H3O+] and [OH-]​

Answer 1

Given:

pH of solution is 3.2

To find:

[H3O+] and [OH-] ?

Calculation:

$pH = 3.2$

$\implies \: - log( {H}^{ + } ) = 3.2$

$\implies \: log( {H}^{ + } ) = - 3.2$

$\implies \: {H}^{ + } = {10}^{ - 3.2} \: M$

$\boxed{H^{+} + H_{2}O\rightarrow H_{3}O^{+}}$

So, 1 M [H+] gives 1 M [H3O+], so 10^(-3.2) M [H+] will give:

$\implies \: H_{3}O^{+} = {10}^{ - 3.2} \: M$

_________________

$pH = 3.2$

$\implies \: p(OH) = 14 - 3.2$

$\implies \: p(OH) = 10.8$

$\implies \: - log(OH) = 10.8$

$\implies \: OH = {10}^{ - 10.8} \: M$

Hope It Helps.

Answer 2

About the question which you have asked to shwetha yesterday

W = μRT log(Pi/Pf)

W = 10 × 10³× 8.3 × 300 × log to the base e (8/4)

By solving this we get

W = 1.728 × 10^7 J

Amount of energy absorbed = Work done = 1.728 × 10^7

Alli 10 × 10³ yake andre thagondirodu 10 kg antha kottiddare alva adakke.. If instead of kg they had given mol means we should not take 10³